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Q.

Two independent  harmonic oscillators of equal mass are oscillating about the origin with angular frequencies ω1  and  ω2 and have total energies E1  and E2 ,respectively. The variations of their momenta P  with positions  x are shown in the figures. If ab=n2 and  aR=n, then the correct equation(s) is (are)

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a

E1ω1=E2ω2

b

ω2ω1=n2

c

ω1ω2=n2

d

E1ω1=E2ω2

answer is B.

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Detailed Solution

1st Particle:  P = 0 at x = a ⇒ ‘a’ is the amplitude of Oscillation 'A1' .  At x = 0 , P = b (at mean position) ⇒mvmax=b .  ⇒vmax=bm----(1) E1=12mvmax2=m2bm2=b22m∵vmax=bm Also, vmax=A1ω1=bm​⇒ω1=bma=1mn2∵a=bn2−−−−−2 2nd Particle:P = 0 at x = R  ⇒A2=RAt x = 0,  P=Rmvmax=R​⇒vmax=Rm−−−−3E2=12mvmax2=m2Rm2=R22m∵vmax=Rm----(4)  Also, vmax=A2ω2=Rm​⇒ω2=RmR=1m−−−−5 (A) E1ω1=b22m×1mn2​E2ω2=R22m×1m​We can see, E1ω1≠E2ω2(B)  ω2ω1=1/m1/mn2=n2(C)  ω1ω2=1mn2×1m=1m2n2C is not correct(D) E1ω1=b2/2m1/mn2=b2n22=a22n2=R22∵a=bn2 & aR=n  E2ω2=R2/2m1/m=R22⇒E1ω1=E2ω2  (D) is correct
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Two independent  harmonic oscillators of equal mass are oscillating about the origin with angular frequencies ω1  and  ω2 and have total energies E1  and E2 ,respectively. The variations of their momenta P  with positions  x are shown in the figures. If ab=n2 and  aR=n, then the correct equation(s) is (are)